Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $x = \dfrac{t + 5}{t - 2} \div \dfrac{4t + 16}{t^2 + 2t - 8} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $x = \dfrac{t + 5}{t - 2} \times \dfrac{t^2 + 2t - 8}{4t + 16} $ First factor the quadratic. $x = \dfrac{t + 5}{t - 2} \times \dfrac{(t - 2)(t + 4)}{4t + 16} $ Then factor out any other terms. $x = \dfrac{t + 5}{t - 2} \times \dfrac{(t - 2)(t + 4)}{4(t + 4)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ (t + 5) \times (t - 2)(t + 4) } { (t - 2) \times 4(t + 4) } $ $x = \dfrac{ (t + 5)(t - 2)(t + 4)}{ 4(t - 2)(t + 4)} $ Notice that $(t + 4)$ and $(t - 2)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ (t + 5)\cancel{(t - 2)}(t + 4)}{ 4\cancel{(t - 2)}(t + 4)} $ We are dividing by $t - 2$ , so $t - 2 \neq 0$ Therefore, $t \neq 2$ $x = \dfrac{ (t + 5)\cancel{(t - 2)}\cancel{(t + 4)}}{ 4\cancel{(t - 2)}\cancel{(t + 4)}} $ We are dividing by $t + 4$ , so $t + 4 \neq 0$ Therefore, $t \neq -4$ $x = \dfrac{t + 5}{4} ; \space t \neq 2 ; \space t \neq -4 $